Abstract

We discuss the existence of subharmonic solutions for nonautonomous second order differential equations with singular nonlinearities. Simple sufficient conditions are provided enable us to obtain infinitely many distinct subharmonic solutions. Our approach is based on a variational method, in particular the saddle point theorem.

1. Introduction and Main Result

In this paper we discuss the problem of the existence of infinitely many subharmonic solutions for nonautonomous second order differential equations with singular nonlinearities of the form
where is continuous, is -periodic, in its first argument with , and presents a singularity with respect to its second argument. Here by a subharmonic solution we mean a -periodic solution for any integer if is the minimal period. When the solution is not -periodic we call it a true subharmonic. It was pointed out in [1] that singular differential equations of the form (1.1) appear in the description of many phenomena in the applied sciences, such as the Brillouin focusing system and nonlinear elasticity. Several authors have investigated the problem of existence of periodic solutions for second order differential equations with singular nonlinearities (see [2–4] and the references therein). Topological and variational methods are the two main techniques that have been developed for the study of (1.1). We refer the interested reader to the paper [1] for details and references on the topological methods. In this work we shall rely on the saddle point theorem, see [5, 6], to prove our main result. We use the truncation techniques introduced in [7] to modify our problem to one without singularities. We assume that the nonlinearity is monotone with respect to its first variable . When is increasing, our result generalizes the result in [8]. We can obtain the same result by considering the monotonicity of the potential function instead of the field . For more results on the subject and different techniques one can consult the papers [9–12]. We should point out some related recent articles, for instance [13, 14].

Throughout this paper we shall use the following notations. Let is the classical Lebesgue space of functions such that is integrable, and for we define its norm by

Let . For and we let and for we define its norm by

endowed with the norm is a reflexive Banach space. Also , orthogonal decomposition, where is the subspace of constant functions in and denotes the subspace of functions in with mean value zero; so that can be written as with and .

We shall assume that is a locally integrable -periodic function. We denote the mean value of by . It follows that , and are bounded. Moreover, since is -periodic, we have .

Let be an antiderivative of defined for all and for all .

We introduce the following assumptions on the nonlinearity.(H1) is continuous, for all , and such that() is monotone for each fixed in ,(), uniformly in ,() for all .(H2)(), uniformly in .().

Theorem 1.1. Assume that is a locally integrable -periodic function. If (H1) and (H2) are satisfied then (1.1) has a sequence of -periodic solutions whose amplitudes and minimal periods tend to infinity. In particular, if is the minimal period of and of with respect to, (1.1) admits solutions with minimal periods for every sufficiently Large Integer .

2.1. Modification of the Problem

Define the truncation function , by
Note that condition (H1) implies that is continuous with respect to and -periodic with respect to its first variable .

Lemma 2.1. Assume (H1) and (H2) (ii) are satisfied. Then there exists such that for every

Proof. First, it follows from (H1)(ii) that for any , there is such that for every we have , uniformly in .In particular for , there is such that for every it holds
Choose such that . Then for every , we have . Therefore, condition (H1)(ii) implies that there exists such that for every it holds , uniformly in .Next, condition (H2)(ii) implies that for any , sufficiently large, there is , large enough, such that for every we have
Hence, there exists , sufficiently large such that and for every it holds
We show that for we have , uniformly in . Assume, on the contrary, that there exists for which , for some . By the continuity of on and (()(iii)), there exists such that . If then , and this is a contradiction. Similarly, if then , and we again arrive at a contradiction. Hence we deduce that if (H1) and (H2)(ii) hold, then there exists such that for any , uniformly in . This completes the proof of Lemma 2.1.

Lemma 2.2. For every positive integer , there exist and with such that for any each -periodic solution of
satisfies
In particular, any -periodic solution of (2.6), with is a solution of (1.1).

Proof. This is essentially Proposition 2.1 in [8]. We shall use some ideas from [8] (see also [15]). Fix , and suppose, on the contrary, that for each integer , there exist and a -periodic solution satisfying
and .Claim 1. Let be as in Lemma 2.1 and let be as above. Then for every there exists such that .Indeed, it follows from (2.8) that
Now, if for all , then Lemma 2.1 implies that , which in turn yields .This contradicts (2.9). On the other hand, if for all , then , so that . This is again a contradiction to (2.9).Claim 2. There exists such that for each integer . To prove the claim notice, there exists such that . If then for any . So, assume that there exists a subsequence of , which we label the same, for which when . So that for large enough. Since , there exists an interval , containing , such that
Equation (2.8) can be written as
Since for all , and , then the second equation in (2.11) is equivalent toLemma 2.1 implies that for all . Then for all and hence is decreasing on . The first equation in (2.11) implies
This yields
Integrating the above inequality over we obtain
Equation (2.15) leads to
Hence
It is clear from (2.17) that when . So that, for large enough, we have
Since , we have . It follows from the first equation in (2.11) that
We see from (2.18) and (2.19) that for large enough
Since is continuous on , then, for large enough, there exists at least one such that . We denote by , the first such . Then
We distinguish two cases.Case 1. is increasing for each fixed .Consider the function defined by
Then
Since , it follows from the first equation in (2.11)
Hence
Since, for all
it follows that for all
Also, Lemma 2.1 implies that for all . Furthermore, the monotonicity of implies that for all . Since is decreasing on and , it follows that
for all . Now, Lemma 2.1 combined with (2.27), (2.28), and the monotonicity of with respect to its first variable shows that
Thus, the function is increasing on . Since ,
Since , it follows that
Notice that
Also, if (see Lemma 2.1). It follows that
Hence
Set
Then
Indeed, since , we have
From (2.37) we deduce
Notice that is bounded, so that
Also, (H2)(i) implies that
Therefore (2.36) holds; that is, . Since and are bounded, it follows that
It is clear that (2.41) contradicts (2.17).Case 2. is decreasing for each fixed .In this case we consider the function defined by
Then
Repeating the same reasoning as in Case 1, we arrive at a contradiction. Therefore, we deduce that there must exist such that for each .Next, we prove that there is such that for every . Assuming that this is not true we will obtain a contradiction.Consider the following sets
It is clear that for , . Also, we cannot have for every , for otherwise we would have
which contradicts the assumption . Hence, for there exists such that . This shows that . The continuity of implies that is open, and so .Define
It follows from (2.9) that . On the other hand
(i)Assume we are integrating positively on all subintervals of .If , then . So that, by Lemma 2.1,
For we have . This means that is bounded uniformly in . Since is continuous it is bounded on .Let
Then
It follows from (2.47), (2.48), and (2.50) that
Claim 3. Proof. Recall that and for each . Then, if we have , and if , we have . In both cases condition (H1)(ii) and the continuity of imply that for every .Since is bounded, then (2.51) implies that
which is a contradiction with (2.9). (ii) If we integrate negatively on all subintervals of we will obtain , which, again, contradicts (2.9). Thus, the proof of Lemma 2.2 is complete.

Remark 2.3. Lemma 2.2 shows that any -periodic solution of (2.6), with is a solution of (1.1), since it satisfies for all and .In the remainder of the paper we shall deal with (2.6), with instead of (1.1). Let be a primitive of defined for all and .

Lemma 2.4. If (H1) and (H2) hold, then and satisfy the following conditions.(L1) is defined and continuous in and -periodic with respect to .(L2), uniformly in .(L3) such that for it holds uniformly in .(L4).

Proof. We shall rely on a variational method and more precisely on the saddle point theorem. Define for each the action functional by is well defined on , weakly lower semicontinuous and continuously differentiable on . Furthermore,
The critical points of are precisely the weak solutions of (2.6), with . First, we show that the functional satisfies the Palais-Smale condition.For this, let be fixed and let be a sequence in such that is bounded and . Then has a convergent subsequence.Suppose, on the contrary, that . Condition (L2) implies that for any , small enough, there is such that
Writing for , we obtain
Since , we have , so that
Now, Hölder’s inequality gives
Since for , it follows from (2.54), (2.58), and (2.59)
Wirtinger’s inequality
combined with the inequality give
This leads to
Indeed, if (2.63) does not hold then there would exist a subsequence of , still denoted the same, which is bounded. Since and are bounded and is chosen arbitrarily small, then (2.62) implies that is bounded. It follows from the inequality
that is bounded, but this contradicts our assumption . Therefore, (2.63) holds. Using Wirtinger’s inequality
in (2.60), we get
It follows from (2.63) that
Using Sobolev’s inequality we obtain
The identity for all and (2.63) imply that
Assume that (the other case can be treated similarly). Then for large enough, uniformly in . By (L3) we have for all
Consequently, for large enough
Since , then for all and for large enough
where for every , and . In particular, if we take in the above inequality we obtain for every
which infer
Now, taking in (2.72) we obtain
Obviously, we have for large enough
Thus, for large enough, (2.75) implies that
Sobolev’s inequality
and Wirtinger’s inequality combined with (2.77) give, for large enough,
So, for large enough, we deduce that
Hence is bounded in . Consequently . Since is bounded, it follows that
is bounded. Hölder’s inequality gives
Since
it follows that
is bounded. But,
Consequently, there exists such that
On the other hand, extending by -periodicity we obtain
Setting for , we get
From (2.75) when uniformly in . By () we have
This is a clear contradiction to (2.86). Therefore, is bounded in , and so it has a convergent subsequence. This shows that satisfies the Palais-Smale condition. Next, we show that has a geometry of a Saddle. For, let then we have and , so that
Proceeding as before, we get an inequality similar to (2.72) by replacing by and by 0,
Since is chosen arbitrary small, we obtain
which shows that is coercive. Hence, admits a bounded minimizing sequence. Furthermore, is weakly lower semicontinuous on , then
For , a constant function, we have if and only if . Then
Extending by -periodicity we obtain
Condition (L4) implies that
Hence, for each is coercive on the space of constant functions. Then for each , there exists , large enough, such that
Thus,
Therefore,
Let be the open interval of centered at 0 and with radius . Since it is clear that
Therefore, we have
with . Thus has a geometry of Saddle. Finally, all conditions of the Saddle point theorem are satisfied. Then for each admits a critical point , which is characterized by
where
Thus for each is a weak -periodic solution of (2.6) with If, furthermore, is assumed continuous, then is a classical solution of (2.6) with . This completes the proof of Lemma 2.5.

Remark 2.6. As a consequence of Lemmas 2.2, 2.5 and the Remark 2.3, we conclude that if () and () are satisfied, then (1.1) admits a sequence of -periodic solutions.

Note that on for each and . This justifies the following definition (see [10, Definition 2.1 page 653]).

Definition 2.7. The level of is defined by when .Every functional admits at least a critical level which is given by .

Note that nondistinct subharmonic solutions have the same level. Then we deduce that in order to find the multiplicity of distinct subharmonic solutions, we have to search the multiple critical levels. The sequence is not always increasing, which means that there exists such that if then .

If for an integer , then is not a level for the -periodic functions and every global minimum of is in fact a subharmonic solution of (2.6).

In our case is not necessary a global minimum for and then even if the condition above is verified, it is still insufficient to deduce the existence of true subharmonic solutions. This is why we prove also that the amplitudes and the minimal periods tend to infinity.

Lemma 2.8. The minimal periods of the solutions of (2.6), with tend to infinity.

Proof. Let be a weak solution of (2.6) with Then is a critical point of . We show that
Let and let be defined for all and for all by
We have and for all . is constant with respect to for all and